1 Answer
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The group $H$ has two elements $\{7, 1\}$ and therefore its index in $G$ is 4. Also note that right cosets coincide with left cosects so that $gH = Hg$ (this comes from the fact that G is abelian and also answers your second question). So e.g. $5 \{7, 1\} = \{11, 5\} = 11 \{7, 1\}$. You'll get 4 cosets like this.

Third question is really the same thing as the first question (thanks to $H$ being normal).

So the Cayley Table for the third part is really the same as the U(24) table OR am I using the 4 cosets as the elements in the table? I have a feeling it is taking 1H, 5H, 13H and 17H and building a Cayley table with those distinct elements.
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spittballzNov 8 '10 at 8:46

I see now that order G = 8, order H = 2 and the index of H in G is now 4 --- so that does work itself out.
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spittballzNov 8 '10 at 8:46

Back to first comment... trying to work that part out. 1H being the identity, all distinct elements times itself gives 1H, but then for example what would 5H*13H become? (11,5)*(19,13) = (19,13,11,5) that result is not 1H,5H,13H or 17H.
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spittballzNov 8 '10 at 8:54

How would I describe this group which is isomorphic to G/H and thus being a nontrivial homomorphic image of U(24)? Using in the description one of the following: cyclic order of..., non-cyclic abelian of order..., or nor-abelian of order...? Not sure what this is asking and how I can answer it.
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spittballzNov 8 '10 at 8:58